To evaluate the limit:

$\lim_{x \to 1} \left( \frac{2}{1-x^2} + \frac{1}{x-1} \right)$

First, simplify each term.

Step 1: Simplify the first term

The expression $\frac{2}{1 - x^2}$ can be factored using the difference of squares formula:

$1 - x^2 = (1 - x)(1 + x)$

Thus,

$\frac{2}{1 - x^2} = \frac{2}{(1 - x)(1 + x)}$

Step 2: Combine the two terms

Now we rewrite the full expression:

$\frac{2}{(1 - x)(1 + x)} + \frac{1}{x - 1}$

Since $x - 1 = -(1 - x)$, the second term becomes:

$\frac{1}{x - 1} = -\frac{1}{1 - x}$

Now the expression becomes:

$\frac{2}{(1 - x)(1 + x)} - \frac{1}{1 - x}$

Step 3: Find a common denominator

The common denominator for the two terms is $(1 - x)(1 + x)$. So we rewrite the second term:

$\frac{-1}{1 - x} = \frac{-1(1 + x)}{(1 - x)(1 + x)} = \frac{-(1 + x)}{(1 - x)(1 + x)}$

Now, combine both terms:

$\frac{2 - (1 + x)}{(1 - x)(1 + x)} = \frac{2 - 1 - x}{(1 - x)(1 + x)} = \frac{1 - x}{(1 - x)(1 + x)}$

Step 4: Simplify the expression

The numerator and one factor of the denominator cancel out:

$\frac{1 - x}{(1 - x)(1 + x)} = \frac{1}{1 + x}$

Step 5: Evaluate the limit

Now, take the limit as $x \to 1$:

$\lim_{x \to 1} \frac{1}{1 + x} = \frac{1}{1 + 1} = \frac{1}{2}$

Thus, the value of the limit is:

$\boxed{\frac{1}{2}}$

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Would you like further details or have any questions?
Here are five related questions you might find interesting:

1. What happens if we try to evaluate this limit without factoring?
2. How does L'Hôpital's Rule apply to limits involving indeterminate forms?
3. How can we evaluate limits at infinity for similar expressions?
4. Can we use series expansions to solve this limit?
5. How do we deal with limits involving trigonometric identities?

Tip: Always check if you can factor expressions to simplify complex limits!